Thermochemistry
Thermochemistry is the subject that deals with the heats involved in chemical reactions.
A typical chemical reaction might have a form similar to the following hypothetical
chemical reaction:
a A + b B  c C + d D.
(1)
(In Equation (1) the upper case letters stand for elements or compounds and the lower
case letters stand for small whole numbers which balance the reaction. You would read
this as saying, "a moles of A reacts with b moles of B to give c moles of C and d moles of
D.")
A chemical reaction is a process just like any other thermodynamic process. It has an
initial state (the reactants) and a final state (the products). We can calculate the changes
in internal energy, enthalpy, and so on for the reaction. For example,
U = U products  U reactants
and
H = H products  H reactants.
One thing is sometimes not made very clear. "Reactants" and "products" in these
equations means that the reactants and products are separated, isolated, and pure.
Furthermore, the reactants and products are all at the same temperature and pressure. So,
for example, the H above is the enthalpy of c moles of C (isolated and pure in its own
container at temperature, T, and pressure, p) plus the enthalpy of d moles of D
(isolated and pure in its own container at temperature, T, and pressure, p) minus the
enthalpy of a moles of A (isolated and pure in its own container at temperature, T, and
pressure, p) minus the enthalpy of b moles of B (isolated and pure in its own container at
temperature, T, and pressure, p).
From time to time we will add a superscript o to H or U to indicate that reactants and
products are in their "standard states." That is, they are in their most stable state at T and
p. For example, the standard state of water at 25oC and 1 atm pressure is liquid water.
If our reaction takes place at constant V, as in a bomb calorimeter, dV =0 and
UV = qV.
If the reaction takes place at constant p, as in open to atmospheric pressure, dp = 0 and
Hp = qp.
If Hp < 0 we say that the reaction is exothermic. That is, the system gave heat to the
surroundings. On the other hand, if Hp > 0 we say that the reaction is endothermic. The
system absorbed heat from the surroundings.
From the definition of enthalpy we find that
H = U + (pV), (2)
where
(pV) = (pV) products  (pV) reactants.
For liquids and solids (pV) is quite small. (pV) is not necessarily small for gases, but
we can get a reasonable estimate for this quantity by approximating the gases as ideal.
Then
(pV)gas  (pV)gas products  (pV)gas reactants  n gas products RT  n gas reactants RT = RTn gas,
where n gas is the difference in the number of moles of gaseous products and reactants.
Using this approximation we get
H = U + RTn gas. (3)
However, we have to be careful how we understand this equation because the conditions
of the reaction must be the same on both sides of the equation. Since H is the heat we
measure if the reaction is run and constant pressure (Hp = qp) and U is the heat we
measure if the reaction is run at constant volume (UV = qV), it is tempting (and common)
to write Equation (3) as
qp = qV + RTn gas. (4)
However, Equation (4) cannot be rigorously true since the q's refer to different
conditions. We can get an indication whether or not Equation (4) is a good
approximation with the following example:
Consider the reaction.
2 C(s) + O2(g)  2 CO(g),
(5)
run at dV = 0.
(pV)  (pV)gas  RTn gas = RT.
The heat measured is qV. Now let us find a way to measure qp Consider the following
two steps
2 C(s) + O2(g)

2 CO(g)

2 CO(g)
(6)
p1, V1, T,
qV
p2, V1, T,
H2
p1, V2, T.
The first step is the constant volume reaction we had before with UV = qV. Notice that
the pressure increases. The second step takes the product of the constant volume reaction
and reduces the pressure back to the original pressure. We call the heat for this step H2.
So it is rigorously true that
Hp = HV + H2 = qV + RT + H2.
(7)
However, the H2 term is the enthalpy for the expansion of a gas at constant temperature.
If the gas is ideal this term is zero. For real gases this term would be very small, so we
make a negligible error is neglecting it. So to pretty good approximation we can use the
equation,
qp = qV + RTn gas. (4)
Hess' Law
Hess' law states that if you add or subtract chemical reaction equations you can (must)
add or subtract their corresponding H's or U's to get H or U for the overall reaction.
For example, if we add the reaction
a A + b B  c C + d D.
H1, U1
to the reaction
eE + fF  gG + hH
H2, U2
to get
a A + b B + e E + f F  c C + d D +  g G + h H,
Then, for the overall reaction
H = H1 + H2 and U = U1 + U2.
The great utility of Hess' law is that we don't have to tabulate H for every possible
reaction. We can get H for a particular reaction by adding and subtracting H's for a
much smaller set of reactions, called formation reactions. We define fH o for a
compound to be the enthalpy of the reaction
pure, isolated elements, in their standard (most stable) states
 one mole of compound in its standard state.
The heat of formation of liquid water is defined as H o for the reaction,
H2(g) + 1/2 O2(g)  H2O(l).
By definition, then, the heat of formation for an element in its most stable state is zero.
To obtain H o for the hypothetical reaction, Equation (1), we add and subtract the
appropriate heats of formations,
rH o = c fHCo + d fHDo  a fHAo  b fHBo. (8)
For example, H o for the reaction,
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l), (9)
is given by
o
o
 r H o   f H CO
 2  f H Ho 2O( l )   f H CH
2(g)
4(g)
 1 mol  (393.51
kJ
kJ
kJ
)  2 mol  ( 285.83
)  1 mol  ( 74.81
)
mol
mol
mol
 890.36 kJ.
(We don't always write out that coefficients which balance the reaction have units, but we
have done so here to make it clear that these numbers have units. This will become an
issue later.)
H at Other Temperatures
Tables of heats of formation usually give data for reactions at 25oC. We frequently need
to know the heat of a reaction at a temperature other than 25oC. If we know the heat
capacities at constant pressure we can calculate the heat of reaction at a temperature other
than 25oC. We use the following chain of reasoning. We know that,
 H 

  Cp
 T  p
then
 H o 
o

  Cp
 T  p
(10a, b, c)
and
 H o 
o

  C p ,
 T  p
where Cpo is defined for our hypothetical chemical reaction, Equation (1) as,
C po  c C poC  d C poD  a C poA  b C poB . (11)
Prepare Equation 10c for integration as
d H o  C po dT , (12)
and integrate,
T2
H o (T2 )  H o (T1 )   C po dT . (13)
T1
For very accurate work we will have to use the temperature dependent heat capacities in
Equations (11) and (13), but more often than not we can regard the heat capacities as
approximately constant over the temperature range, so that Equation (13) becomes,
H o (T2 )  H o (T1 )  C po (T2  T1 ). (14)
As an example, let's calculate the heat of reaction for the reaction in Equation (9) at 95oC.
Cpo for the reaction in Equation (9) is given by,
 rC po  C poCO2 ( g )  2 C poH 2O( l )  C poCH 4 ( g )  2 C poO 2 ( g )
J
J
)  2 mol  (75.291
)
K mol
K mol
(15)
J
J
1 mol  (35.31
)  2 mol  (29.355
)
K mol
K mol
 1 mol  (37.11
 93.67
J
.
K
Then, Equation (14) gives,
H o (95o C)  H o (25o C)  C po (368.15  298.15)
 890.36 kJ  93.67
J
1 kJ
 70 K
(16)
K
1000 J
=  883.80 kJ
(Note: There is another way to do this problem. Calculate H o for cooling the reactants
down to 25oC, calculate H o for the reaction at 25oC, calculate H o for heating the
products back up to 95oC, and then add them up. Since H is a state function H is
independent of path. This method will also work if one of the components of the reaction
has a phase change somewhere in the temperature range.
For example, if we let the new temperature be over 100oC we would have to account for
the vaporization of the liquid water product. This is not hard to do, but requires some
extra steps, including also the use of the heat capacity of water vapor. If the upper
temperature in this problem is above 100oC we carry out the reaction in several steps:
Step 1 - Cool the reactants from the upper temperature to 25oC,
Step 2 - Run the reaction at 25oC,
Step 3 - Heat the product CO2 from 25oC to the upper temperature,
Step 4 - Heat the liquid water from 25oC to 100oC,
Step 5 - Vaporize the water at 100oC,
Step 6 - Heat the water vapor from 100oC to the upper temperature.
The heat of reaction at the upper temperature is the sum of the H 's for all six steps.
Again we have taken advantage of the fact that H is independent of path.)
H as Making and Breaking Chemical Bonds
Breaking a chemical bond is an endothermic process. That is, you must put energy into
the system to break the bond.
Forming a chemical bond is an exothermic process. The energy released in forming the
bond goes into the surroundings.
We can make an estimate of the H for a chemical reaction by adding the bond energies
of all the bonds broken and subtracting the bond energies of all the bonds formed.
H  BE bonds broken  BE bonds formed.
Let's try it on the gas phase reaction,
N2 + 3 H2  2 NH3.
(There are tables of bond energies in physical chemistry texts and in data handbooks.
From the tables we find the following bond energies:
NN 945 kJ
HH 436 kJ
NH 388 kJ.
In the reaction we break 1 NN bond and 3 HH bonds. We form 6 NH bonds. The
approximate H is then,
H  1  945 + 3  436  5  388 =  75 kJ.
This can be compared to the actual value of  92 kJ. The method is not super accurate,
but it gives a ball-park answer and might be useful in cases where other data are not
available. Further, however it demonstrates graphically that the heat of a reaction is
related to the making and breaking of chemical bonds.
Heats of Formation of Ions in Water Solution
When you look in a table of heats of formation you find values listed for ions in water
solutions. That is, you will find an entry for species such as Na+(aq). It is fair to ask
where these numbers come from. We know that in equilibrium chemistry it is impossible
to prepare Na+(aq) ions in solution all by themselves. It is possible to prepare a solution
that has both Na+(aq) and Cl(aq) ions, but not a solution that has only ions of one
charge.
The heats of formation of solutions of soluble ionic compounds can be measure. That is,
we can measure the heat of formation of HCl(aq). The heat of formation of HCl(aq) is
defined as H o for the reaction,
1/2 H2(g) + 1/2 Cl2(g)  HCl(aq). (17)
Since ionic compounds in solution are completely dissociated, it must be true that
o
o
o
 f H HCl(
aq )   f H H  ( aq )   f H Cl ( aq ) . (18)
We cannot know the heat of formation of either of the ions in solution, but we do know
their sum.
By convention we arbitrarily set the heat of formation of the H+(aq) ion equal to zero.
That is,
 f H Ho  ( aq )  0. (19)
Then the heats of formation of all other aqueous ions can be determined relative to the
heat of formation of the H+(aq) ion. As a start, we see that
o
o
 f H Clo  ( aq )   f H HCl(
aq )   f H H  ( aq )
o
  f H HCl(
aq )  0
(20a, b, c)
o
  f H HCl(
aq )
This convention allows us to build up a table of heats of formation of aqueous ions.
For example, from the measured value of the heat of formation of NaCl(aq) and the
knowledge that
o
o
o
 f H NaCl(
aq )   f H Na  ( aq )   f H Cl ( aq ) . (21)
we find that
o
o
o
 f H Na
  f H NaCl(

aq )   f H Cl ( aq )
( aq )
 f H
o
NaCl( aq )
 f H
o
HCl( aq )
(22a, b)
.
Continuing these procedures we can define the heats of formations of other aqueous ions,
o
o
 f H Bro  ( aq )   f H HBr(
aq )   f H H  ( aq )
o
  f H HBr(
aq )  0
(23a, b, c)
o
  f H HBr(
aq ) .
We now have enough data in our table to calculate the heat of formation of aqueous NaBr
without measuring it.,
o
o
o
 f H NaBr(
aq )   f H Na  ( aq )   f H Br  ( aq )
o
o
o
  f H NaCl(
aq )   f H HCl( aq )   f H HBr( aq ) .
(24a, b)
If the heat of formation of H+(aq) were set to some value other than zero, it would have
cancelled out of Equation (24b). In this manner an entire table of heats of formation of
aqueous ions can be built will all values relative to the heat of formation of the H+(aq)
ion.